For what vectors will $I-uv^*$ be singular? Let $u,v$ be $1\times m$ vectors. For what vectors $u$ and $v$ will $A:=I-uv^*$ be singular? And if it is singular, what is its nullity?
My approach is that the rank of $uv^*$ is $1$, so $m-1$ eigenvalues are zero and one of the eigenvalues of $uv^*$ must be $1$. This is because $(I-uv^*)x=(1-\lambda')x$, $x\ne 0$. Now, we can represent the characteristic polynomial of $uv^*$ as $\lambda'^{(m-1)}(\lambda'-u_{ij} \overline{v}_{ji})$ for some $i,j$. Does this imply that $u,v$ must satisfy the condition that $u$ and $v$ must have one entry $u_{ij}$ and $v_{ij}$ such that $u_{ij}v_{ji}=1$, that is $u_{ij}$ and $v_{ji}$ are roots of unity, and are equal?
 A: $$\DeclareMathOperator{vspan}{span}$$I don't think your expression for the characteristic polynomial is correct (since $u, v$ are vectors, they should only have on index).  I suspect that if you fix the expression, you would get the correct result (which includes more vectors than what you have currently).
However, there is an easier way to simplify the problem.  As you noticed, the problem is equivalent to finding when $uv^*$ has an eigenvalue of $1$.  Notice that
$$(uv^*)w = (v^*w)u$$
so the eigenvector of $uv^*$ must be $u$.  On the other hand, 
$$(uv^*)u = (v^*u)u$$
so the eigenvalue of $u$ is $v^*u$.  Thus, the condition is that $v^*u = 1$.

copper.hat in the comments to the question mentions the Sherman-Morrison formula.  This formula gives the expression for $(A + uv^*)^{-1}$ in terms of $A^{-1}$, $u$, and $v$.  In particular, it states that
$$(A + uv^*)^{-1} = A^{-1} - \frac{A^{-1}u v^* A^{-1}}{1 + v^* A^{-1} u}.$$
Taking $A = I$, we see that we can compute $(I + uv^*)^{-1}$ whenever $1 + v^*u \neq 0$, confirming our prior calculation.
A: Let $(\lambda,w)$ be an eigenpair of $A$,then
$$Aw = (I - uv^*)w = \lambda w$$i.e.
$$w - (v^*w)u = \lambda w$$
Let $\lambda = 0$, then
$$w = (v^*w)u \tag{1}$$
This means that an eigenvector in the null space should be along $u$, hence $w = \alpha u$ and therefore $(1)$ becomes
$$(v^*\alpha u)u = \alpha u $$
which gives
$$(v^*u)u = u $$

which means that $v^*u = 1$. This is the only condition you need so that $I - uv^*$ is singular. Also, and as you argue, this will mean that under the condition that $v^*u = 1$, the only eigenvector in the null-space is $u$.

